Problem: In rectangle $ABCD$, angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$, where $E$ is on $\overline{AB}$, $F$ is on $\overline{AD}$, $BE=6$, and $AF=2$. Find the area of $ABCD$.

[asy]
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); dotfactor=4;
real length = 2 * (6*sqrt(3) - 2), width = 6*sqrt(3);
draw(origin--(length,0)--(length,width)--(0,width)--cycle);
draw((length,width)--(0,2)^^(length,width)--(length - 6,0));
dot("$A$",origin,SW); dot("$B$",(length,0),SE); dot("$C$",(length,width),NE); dot("$D$",(0,width),NW); dot("$F$",(0,2),W); dot("$E$",(length - 6,0),S);
[/asy]
Answer: From $30^\circ$-$60^\circ$-$90^\circ$ triangle $CEB$, we have $BC=6\sqrt{3}$. Therefore, $FD=AD-AF=6\sqrt{3}-2$. In the $30^\circ$-$60^\circ$-$90^\circ$ triangle $CFD$, $CD=FD\sqrt{3}=18-2\sqrt{3}$. The area of rectangle $ABCD$ is $$(BC)(CD)=\left(6\sqrt{3}\right)\left(18-2\sqrt{3}\right)=
\boxed{108\sqrt{3}-36}.$$